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I was reading this post and I got really curious about that expression. I mean, I'd like to compute the limit: $$ \lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}} $$

What I tried: following the advice, I took logarithms.

$$ \lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}} \Rightarrow \ln\left ( \lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}} \right ) = n^2\cdot\ln\left ( \frac{1}{n} \right )\cdot\frac{1}{n}\cdot\sum_{k=1}^{n} \ln \left ( 1 + \frac{n}{k} \right ) $$

As $ \lim_{n\to\infty} \frac{1}{n}\cdot\sum_{k=1}^{n} \ln \left ( 1 + \frac{n}{k} \right ) = \int_{1}^{2} \ln(1+x) dx $, then:

$$ \lim_{n\to\infty} n^2\cdot\ln\left ( \frac{1}{n} \right )\cdot\underbrace{\int_{1}^{2} \ln(1+x) dx}_{\sim 0.9 >0} = -\infty $$

Am I right?

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  • $\begingroup$ Where did you miss your $x$ in equation? Or there are must be another variable? $\endgroup$ – itdxer Jul 1 '15 at 17:50
  • $\begingroup$ Do you mean $n$ instead of $x$ $\endgroup$ – Elaqqad Jul 1 '15 at 17:58
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    $\begingroup$ Yes, you are correct. You need only evaluate $e^{\lim_{n\to \infty}\log (\cdots )}=0$ and you are done! $\endgroup$ – Mark Viola Jul 1 '15 at 20:27
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Your reasoning is correct and the limit is $0$, in order to see it more clearly you can use the inequality: $$ n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}}\leq \frac{\prod_{k=1}^n (2n)^{\frac{1}{n}}}{n^{n^2}}=\frac{2n}{n^{n^2}} $$ and here you can see why ...

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